The volume under the surface $z (x, y) = x + y$ and above the triangle in the $x - y$ plane defined by $0 \leq y \leq x a n d 0 \leq x \leq 12$ is
864

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Solution

The correct option is A 864
Volume $= \int_{x = 0}^{12} \int_{y = 0}^{x} \int_{z = 0}^{x + y} (1) d z d y d x$
$= \int_{x = 0}^{12} \int_{y = 0}^{x} (x + y) d y d x = 864$

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A triangle in the xy plane is bounded by the straight lines 3 x = 4 y, y = 0 and x = 4 . The volume above the triangle and under the plane x + y + z = 8 is

x y -

2 x = 3 y, y = 0

x = 3

x + y + z = 6

f (x, y)

(x, y) ϵ [0, 1] \times [0, 1]

x > y^{2}

y > x^{2}

f (x, y)

x - 2 y^{2} \geq 0, 1 - x - | y | \geq 0

y + \sqrt{3} x = 6, y - \sqrt{3} x = 6,

y = 0,

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